### We will focus on a few simple problems where we will manipulate the equations for relativistic energy and momentum.

*This could be seen as a second-year university-level post.*

Einstein had shown that the Lorentz transformations were the correct way to switch between the coordinate systems of different frames of reference [1]. He also taught us that Newton’s laws weren’t at all proper relativistic laws. For instance, Newtonian momentum $ \mathbf{p} = m \mathbf{v} $, and energy $ E = mv^2 / 2 $ were not at all accurate at speeds approaching that of light.

Instead, we have all come to learn that the relativistic momentum is written as

\begin{equation} \label{eq:relativistic momentum} \mathbf{p} = \frac{m \mathbf{v}}{\sqrt{1 – \dfrac{v^2}{c^2}}}. \end{equation}

And that the correct relativistic expression for total energy is

\begin{equation} \label{eq:relativistic energy} E_{\text{tot}} = \frac{mc^2}{\sqrt{1 – \dfrac{v^2}{c^2}}}. \end{equation}

We will solve the following problem set:

- Prove, for a particle travelling at $ c $, that the magnitude of the relativistic energy is given by $ E = pc $.
- Show that the energy-momentum relation for a particle with
*any*mass $ m $ travelling at*any*speed $ v $ is correct and do mind it is*not*the famous $ E = mc^2 $ we are referring to. Use the correct one, if you please. - Given that the mass of a proton is $ m_p $, calculate its exact speed when its relativistic translational kinetic energy (which is the relativistic total energy minus its relativistic mass energy) is four times its relativistic mass energy.

**Problem I**

Since $ E $ is expressed in terms of $ p $, we need to rewrite Eq. $ \eqref{eq:relativistic momentum} $ by solving for $ m $:

\[ m = \frac{p \sqrt{1 – \dfrac{v^2}{c^2}}}{v}. \]

Note, we do not use the vector quantities, just the magnitudes. We can now proceed to substitute this into Eq. $ \eqref{eq:relativistic energy} $:

\[ E_{\text{tot}} = \frac{\left(\dfrac{p \sqrt{1 – \dfrac{v^2}{c^2}}}{v}\right)c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}. \]

This reduces to

\begin{align}

E_{\text{tot}} &= \frac{pc^2 \sqrt{1 – \dfrac{v^2}{c^2}}}{v \sqrt{1 – \dfrac{v^2}{c^2}}}, \\

\therefore E_{\text{tot}} &= \frac{pc^2}{v}. \label{eq:E=pc^2/v}

\end{align}

As we are dealing with a particle travelling at speed $ c $, we know $ v = c $, rendering Eq. $ \eqref{eq:E=pc^2/v} $ to

\begin{align}

E_{\text{tot}} &= \frac{pc^2}{c}, \\

\therefore E_{\text{tot}} &= pc.

\end{align}

**Problem II**

The energy-momentum relation is

\[ E^2_{\text{tot}} = p^2c^2 + m^2c^4. \]

Substituting Eqs. $ \eqref{eq:relativistic momentum} $ and $ \eqref{eq:relativistic energy} $, yields

\[ \left(\frac{mc^2}{\sqrt{1 – \dfrac{v^2}{c^2}}}\right)^2 = \left(\frac{m \mathbf{v}}{\sqrt{1 – \dfrac{v^2}{c^2}}}\right)^2c^2 + m^2c^4, \]

which we can continue to work out as follows:

\begin{align*}\left(\frac{mc^2}{\sqrt{1 – \dfrac{v^2}{c^2}}}\right)^2 – \left(\frac{m \mathbf{v}}{\sqrt{1 – \dfrac{v^2}{c^2}}}\right)^2c^2 – m^2c^4 &= 0, \\

\frac{m^2c^4}{1 – \dfrac{v^2}{c^2}} – \frac{m^2v^2c^2}{1 – \dfrac{v^2}{c^2}} – m^2c^4 &= 0, \\

\left(1-\dfrac{v^2}{c^2}\right)\left(\frac{m^2c^4}{1-\dfrac{v^2}{c^2}}\right) \qquad &\qquad \\ – \left(1-\dfrac{v^2}{c^2}\right)\left(\frac{m^2v^2c^2}{1-\dfrac{v^2}{c^2}}\right) &\qquad \\ – \left(1-\dfrac{v^2}{c^2}\right)m^2c^4 &= 0, \\

m^2c^4 – m^2v^2c^2 – m^2c^4 + \frac{m^2v^2c^4}{c^2} &= 0, \\

m^2c^4 – m^2c^4 – m^2v^2c^2 + m^2v^2c^2 &= 0, \\

0 – 0 &= 0.

\end{align*}

Hence, for every value of $ m $, $ p $, and thus $ v $, the relation holds.

**Problem III**

The relativistic (total) energy is

\[ E_{\text{tot}} = E_{\text{trans}} + E_{\text{mass}}. \]

If the relativistic translational kinetic energy is four times the relativistic mass energy, then we can write

\[ E_{\text{trans}} = 4E_{\text{mass}}. \]

In our case, this then yields for the relativistic (total) energy:

\[ E_{\text{tot}} = 4E_{\text{mass}} + E_{\text{mass}} = 5E_{text{mass}}. \]

To calculate the proton’s speed, we then write

\begin{align*}

\frac{m_pc^2}{\sqrt{1 – \dfrac{v^2}{c^2}}} &= 5E_{\text{mass}} = 5m_pc^2, \\

\frac{1}{\sqrt{1 – \dfrac{v^2}{c^2}}} &= 5, \\

\sqrt{1-v^2/c^2} &= \frac{1}{5}, \\

1-\frac{v^2}{c^2} &= \frac{1}{25}, \\

\frac{v^2}{c^2} &= \frac{24}{25}, \\

v^2 &= \frac{24c^2}{25}, \\

\therefore v &= \sqrt{\frac{24c}{25}} = \frac{2\sqrt{6}c}{5},

\end{align*}

which is about $ 0.98c $ rounded to two decimals, which means that the proton zips at about 98% of the speed of light through the fabric of the cosmos.

*Image Hydrogen bubble chamber Fermilab: Proton with 300 GeV energy producing 26 charged particles in the 30 inch hydrogen bubble chamber at Fermilab. Source: **Wikimedia Commons*

*Annalen der Physik*, 322(10), pp. 891–921. doi: 10.1002/andp.19053221004.

This is a repost. Slight errors in the parsing of LaTeX in the original article of 24 December 2018 have been corrected.

*@kjrunia** is reading mathematics and theoretical physics (final year) in England, at The Open University, Walton Hall, Milton Keynes. He also works on coding for the Mars Rover of the university’s Planetary Robotics Team.*